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3 years ago

If its mass is 3.55×105 kg and the net braking force is 4.05×105 n , what is the airplane's acceleration?mastering physics

2 answers:

6 0

This problem illustrate an airplane which is decelerating, since it tells us about a braking force, this type of acceleration is opposite in direction or simply a "deceleration" of the aircraft. The given are mass m=3.55 x 10⁵ Kg. The net braking Force f=-4.05 x 10⁵ N or Newton. The formula for the force is equivalent to F=ma. therefore a=f/m or force divided by the mass. This braking force is acting in opposite direction therefore having a negative sign. The answer will be negative in sign reminding us that it is a form of "deceleration". The unit should be in m/s².

solong [7]3 years ago

5 0

Answer:

$-1.14 m/s^2$

Explanation:

We can find the airplane's acceleration by using Newton's second law of motion, which states that:

$F=ma$

where

F is the net force on the airplane

m is the mass of the airplane

a is its acceleration

The problem says that the force is a braking force, so this means that the airplane is slowing down, therefore the acceleration must be negative and we must put a negative sign in front of the force. Re-arranging the equation, we can find the acceleration:

$a=\frac{F}{m}=\frac{-4.05\cdot 10^5 N}{3.55\cdot 10^5 kg}=-1.14 m/s^2$

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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

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Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos:

$T = \frac{1}{f}$ (1)

Donde $f$ es la frecuencia, en hertz.

( $f = 4\,hz$ )

$T = \frac{1}{4\,hz}$

$T = 0.25\,s$

Ahora determinamos la fuerza aplicada sobre la prenda ( $F$ ), en newtons:

$F = m\cdot a$ (2)

$F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}$ (2b)

Donde:

$m$ - Masa de la prenda, en kilogramos.

$r$ - Radio interior del tambor, en metros.

( $m = 0.32\,kg$ , $r = 0.4\,m$ , $T = 0.25\,s$ )

$F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}$

$F \approx 80.852\,N$

La velocidad lineal de la lavadora es:

$v = \frac{2\pi\cdot r}{T}$ (3)

( $r = 0.4\,m$ , $T = 0.25\,s$ )

$v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}$

$v \approx 10.053\,\frac{m}{s}$

Y la velocidad angular del tambor de la lavadora:

$\omega = \frac{2\pi}{T}$

( $T = 0.25\,s$ )

$\omega = \frac{2\pi}{0.25\,s}$

$\omega \approx 25.133\,\frac{rad}{s}$

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3 years ago

A rock thrown with speed 8.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m bef

frez [133]

Draw a diagram to illustrate the problem as shown below.

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v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
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Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
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Answer: The height is 21.7 m (nearest tenth)

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3 years ago